I am doing some physics problem and in order to solve it I need to solve the following integral:
$$ \int_{0}^{\pi} 2\pi\frac{R^2(z-R\cos\theta)\sin\theta }{\sqrt{(R^2+z^2-2zR\cos\theta)^3}}\,d\theta $$ where $z>R$.
Considering physics of the problem I expect to get $\dfrac{4\pi R^2}{z^2}$ (although that can be wrong). I tried to do substitution $u=R^2+z^2-2zR\cos\theta$, but I did not succeed to get the solution. Also i tried to convert that real integral into complex integral to solve it, but also I did not succeed to get the solution. So my question is how to solve that integral?
Thank you for any help!